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14.Probability
easy
If two events $A$ and $B$ are such that $P\,(A + B) = \frac{5}{6},$ $P\,(AB) = \frac{1}{3}\,$ and $P\,(\bar A) = \frac{1}{2},$ then the events $A$ and $B$ are
A
Independent
B
Mutually exclusive
C
Mutually exclusive and independent
D
None of these
Solution
(a) We have $P(A + B) = P(A) + P(B) – P(AB)$
$ \Rightarrow \frac{5}{6} = \frac{1}{2} + P(B) – \frac{1}{3} $
$\Rightarrow P(B) = \frac{4}{6} = \frac{2}{3}$
Thus, $P(A)\,.\,P(B) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} = P(AB)$
Hence events $A$ and $B$ are independent.
Standard 11
Mathematics
